Percent (%) Composition of CuO
Cu = 1 x 50g - Multiply by one as there is one Cu
O = 1 x 12.5g - Multiply by one as there is one O
CuO = 62.5g
% for Cu = 50g over 62.5 multiplied by 100 = 80%
% for O = 12.5g over 62.5 multiplied by 100 = 20%
Final Answer :
<em>Percent (%) Composition of CuO = </em>80% (Cu) & 20% (O)
This question seems to be an essay question from experiment. Different solution of oxidizing agent will have different strength. Sulfuric acid or H2SO4 is weaker oxidizing agent when compared to nitric acid (HNO3). In this case, if you subtitute the H2SO4 you wouldn't be able to get the same result for the experiment.
<span>a particular kind of matter with uniform properties.
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Answer:
38
Explanation:
Symbol: Sr
Atomic mass: 87.62 u
Electrons per shell: 2,8,18,8,2
Atomic number: 38
Electron configuration: [Kr] 5s2
Van der Waals radius: 255 pm
Valence electrons: two
Answer:
It was done correctly, Your answer is completely correct.
Explanation: